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Let (K) be the pivotal cubic pK(Ω = p : q : r, P = u : v : w) which meets the circumcircle (O) at A, B, C and three other points T1, T2, T3, vertices of the triangle denoted (T). These points are not necessarily all real and two of them can be imaginary conjugates. In the following, we assume that (K) is not a circular cubic, so none of the points T1, T2, T3 is a circular point at infinity. See CL035 for a study of circular pKs. Recall that : • when Ω = X(6), P must not lie on the line at infinity. • when P = X(4), Ω must not lie on the orthic axis. • when Ω ≠ X(6) and P ≠ X(4), Ω (resp. P) must not be the unique pole Ωcirc (resp. pivot Pcirc) such that (K) is circular. See Special Isocubics §4.2.1 for further details.
The isogonal conjugation θ with respect to (T) transforms (K) into another cubic (K') also passing through T1, T2, T3 and meeting (O) again at S1, S2, S3. (K') is a central cubic which is called the adjunct central cubic of (K). Points on (K') • θ maps (O) to the line at infinity hence the infinite points of (K') are the transforms of A, B, C. • θ maps the line at infinity to (O) hence S1, S2, S3 are the transforms of the infinite points of (K). • (K) passes through its pivot P, its isopivot or secondary pivot Q = Ω ÷ P, its tertiary pivot R = P / Q (PCeva conjugate of Q). Hence, (K') must contain their transforms X, Y, Z respectively. It turns out that X is the center of (K') and Y, Z are symmetric about X.
Some examples of pairs {(K), (K')} {K002, K758}, {K003, K1267}, {K004, K004}, {K006, K1362}, {K007, K1327}, {K169, K1361}.
A special case : (K) is an isogonal cubic When Ω = X(6), the pivot P is the orthocenter of (T) hence X must be O for every P. In this case, S1, S2, S3 are the antipodes of T1, T2, T3 on (O). In particular, when P lies on the Euler line, (K) is a cubic of the Euler pencil as in Table 27. In this case, (K) and (K') meet at O, P, T1, T2, T3 and four other points which lie on K800 and on two perpendicular lines secant at the complement cP of P. These points also lie on the rectangular hyperbola with center cP passing through O, X(2574), X(2575), the isogonal conjugate gP of P in ABC, the reflection of P about X(5).
The general case : (K) is not an isogonal cubic When Ω ≠ X(6) and P ≠ X(4), one can find two isogonal pKs with pivots P1 and P2 which meets (L∞) and (O) at the same points as (K) respectively if and only if Ω lies on the line through X(6), Ωcirc, or equivalently, P lies on the line through X(4), Pcirc. In this case, the adjunct central cubic (K') of (K) meets (L∞) at the same points as K004, i.e. at the infinite points of the altitudes of ABC. It meets (O) at T1, T2, T3 and at S1, S2, S3 which are the antipodes of the points (apart A, B, C) where pK(X6, P1) meets (O). The center X of (K') lies on the line passing through Ω ÷ X(4) and the trilinear pole of the line passing through X(6) and Ω. This latter point is on (O). The sidelines of (T) envelope a parabola with focus on the circumcircle whose directrix passes through P2. The conic inscribed in both triangles ABC and (T) is tangent to a fixed line which is also tangent to the parabola.
to be continued... 
